How many mol of sodium hydroxide are required to make 1.35 L of 2.50 mL solution?
1 answer:
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32.8 g of Butane is required and 99.3 g of CO₂ is produced
<u>Explanation:</u>
The above mentioned reaction can be written as,
C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g) where ΔH (rxn)= -2658 kJ
It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.
That is
of butane reacted
Now this moles is converted into mass by multiplying it with its molar mass = 0.564 mol × 58.122 g / mol
= 32.8 g of butane.
Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol
= 99.3 g of CO₂
Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced
The composition of the stack gas are :
= 0.8713
= 0.0202
CO = 0.107
<h3 /><h3>What is a mole fraction?</h3>The ratio of the number of moles of one component of a solution or other mixture to the total number of moles representing all of the components.
Assuming 100 g of the stack gas. Calculate the mass of each species in this sample according to their percentages.
Mass of : 70% of 100 g = 70 g
Mass of : 15% of 100 g = 15 g
Mass of CO : 15% of 100 g = 15 g
Now calculate the number of moles of each species:
Number of moles of :
= 4.3 mole
Number of moles of :
= 0.10 mole
Mass of CO : = 0.53 mole
Now to calculate the mole fraction of each we use the formula:
Mole fraction of :
= 0.8713
Mole fraction of :
= 0.0202
Mole fraction of CO : = 0.107
Hence, composition of the stack gas are:
= 0.8713
= 0.0202
CO = 0.107
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